Question

help help help for a medal


the company discovered that it costs $16 to produce 2 widgets, $18 to produce 4 widgets, and $48to produce 10 widgets. Using the quadratic function, find the cost of producing 6 widgets.

Answer 1

there is actually a way to solve this but it is rather a pain
i would use technology
http://www.wolframalpha.com/input/?i=quadratic+%282%2C16%29%2C%284%2C18%29%2C%2810%2C48%29

Answer 2

looks like your quadratic function is
\[f(x)=\frac{1}{2}x^2-2x+18\]

Answer 3

to answer the question posed, find
\[f(6)\]

Answer 4

This problem looks awfully familiar.

Answer 5

@satellite73 so would i just plug in 6 into the x's?

Answer 6

have you done one similar? @Johnbc

Answer 7

yes that is what you would do

Answer 8

I have watched a similar problem be performed but in a different manner

Answer 9

you have to solve a three by three system of equations to do this, and it is annoying

Answer 10

oh okay well thanks for the help website helped out great @satellite73

Answer 11

yw
note the syntax

Answer 12

seems really difficult ... @Johnbc

Answer 13

Yeah that is exactly right, it turns out to be 3 different values we are looking for (a, b, and c) and there will be 3 different equations so it will be system of equations but I like and trust satellite73's method.

Answer 14

Quite simple but tedious.

Answer 15

would you be able to explain or is it to late ? @Johnbc

Answer 16

\[ax^2+bx+c\]
\[a\times 2^2+b\times 2+c=16\] or
\[4a+2b+c=16\] is equation one

Answer 17

similarly get two more equations
solve the system

Answer 18

Of course,

As a quadratic function
\[Cost = ax^2+bx+c\]
Where X is the amount of widgets.

So for each
\[16 = a(2)^2+b(2)+c\]
\[18 = a(4)^2+b(4)+c\]
\[48 = a(10)^2+b(10)+c\]

Answer 19

You can use one the equations to solve for the values of a, b, and c as system.

Answer 20

so would you plug in 6 ?
@Johnbc ahhh im so confussed ):

Answer 21

In the method satellite showed, you would plug in 6

Answer 22

thank you so much i got it my answer was 24 thank you :) @Johnbc

Answer 23

My pleasure

Answer 24

by any chance would you be able to figure out the quadratic function ? like satellite73

Answer 25

I am going to solve it using the other method so you can refer to it next time if you get stuck again.

Using
\[16= 4a+2b+c\]

We subtract it from the other 2 equations so
\[18=16a+4b+c\]\[-(16 = 4a+2b+c)\]_________________________
2 = 12a + 2b

And\[48 = 100a+10b+c\]\[-(16=4a+2b+c)\]______________________________
32 = 96a + 8b

Answer 26

Using both our new equations,

I multiply my smaller equation by 8 to get
16 = 96a + 16b

Because we want to eliminate another term as we did for the "c" terms so this eliminates the "a",

\[32 = 96a + 8b\]\[-(16= 96a +16b)\]____________________
16 = 0 - 8b
b = -2

Using b =-2 back into 2 = 12a +8b

\[2 = 12a +2(-2) \]\[2 = 12a-4\]\[6 = 12a\]\[a = \frac{ 6 }{ 12 } = \frac{ 1 }{ 2 } \]

Answer 27

We know have b = -2 and a = 1/2 so using our original equation
\[16= 4a +2b +c\] we can solve for c

\[16 = 4(\frac{ 1 }{ 2 })+2(-2) +c\]
\[16 = 2 -4 +c\]
\[18 = c\]

Answer 28

Now that we have a. b, and c we can find our quadratic function as

\[cost = \frac{ 1 }{ 2 }x^2-2x+18\]

Answer 29

ommmmgg thank you so muchh !

Answer 30

My pleasure

Answer 31

It is exactly as Satellite73 said it would be!
& Now you have the steps for reference.